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I need to pass high current on some part of my circuit. I used an online PCB track width calculator to see that required track width is about 5mm and minimum clearance is 1mm, which makes it about 7mm width at total just for one track. I need several of these high current carrying tracks on my PCB which will consume too much space to afford.

I am thinking of soldering copper wires on the top side of the PCB which will be parallel to the thin and symbolical tracks on the bottom side. But I would like to know if there is a more professional way of overcoming this problem.

hkBattousai
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  • stevenvh's and Olin Lathrop's answers are very-much direct. The same cross-sectional area, with increased-thickness or height, would take lesser width. – Always Confused Jul 04 '16 at 09:13
  • But what would be the board's characteristics? how much thick or thin board should be chosen? – Always Confused Jul 04 '16 at 09:15
  • Is there a possibility of bending-up of the board on heating due-to a "bimetallic-strip"-effect? – Always Confused Jul 04 '16 at 09:17
  • Also another-point, high-current also means requirement of higher-voltage. So, there is higher-chance of **Sparking, Leakage, Short-circuit,** etc. So, they are need to be prevented. – Always Confused Jul 04 '16 at 09:33
  • @AlwaysConfused how is there a *requirement* of higher voltage? If OP is running 5V @ 12A (which would call for ~4.62mm trace @ 2oz), it's 5V... there's no sudden need to increase the voltage. If OP has space to do so at either end, they *could* boost the voltage at the source and then downconvert at the destination to *reduce* current requirements... but high current does not, in itself, necessitate high voltage in *any* meaningful way. – Doktor J Apr 16 '19 at 16:03

7 Answers7

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I haven't seen anybody else mention temperature.

Perhaps you left the default 10 degree rise in the online calculator?

That's pretty conservative. A 20 degree rise isn't that bad in a lot of situations.

And if you aren't running the highest current continuously, it's quite possible even a higher temp rise would be acceptable, since it will have time to cool down between cycles.

Brock R.
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    This is the right answer of this question. How strange no one voted it up. – johnfound Oct 13 '13 at 08:24
  • related: http://electronics.stackexchange.com/questions/243973/if-i-pass-electricity-through-heating-coils-will-it-make-heat/243987#243987 – Always Confused Jul 04 '16 at 09:26
  • **from Joule's law ; indeed decreased resistance (more-conductance) give-rise to more-heating**. H= (const.) * (I^2) * R * t . . . . . ( t is time here) => H= (const.) * (I^2) * (V/I) * t => H= (const.) * I * V * t. ( => H varies directly in-proportion with current I , when you are using a single piece of Conductor). The impact of varying R could be understood only when 2 or more heater will-be kept in series, so the same current will flow through all these resistors. Then the heater with highest resistance ("tightest"-one) produce more heat in comparison to the other heaters. – Always Confused Jul 04 '16 at 09:28
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High-current PCB bus bars are available from several suppliers, such as:

http://www.espbus.com

and are an ideal solution. A quick search for "PCB bus bars" will yield a number of suppliers.

Doktor J
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Leon Heller
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    Second link is dead. – Bradman175 Aug 14 '16 at 01:29
  • -1, all the links are dead – duedl0r Aug 05 '17 at 22:30
  • One link fixed. – Leon Heller Aug 06 '17 at 07:15
  • A search for "busbar" (no space) on the standard supplier sites, such as Digikey (https://www.digikey.com/products/en?keywords=busbar ) gives some off-the-shelf products. High volume PCBs use custom busbars (from the suppliers you'll find on Google), but since you are creating the PCB you could design around off-the-shelf parts: busbar rods, busbar terminals, ... – Casey Apr 22 '20 at 17:48
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The first answer would be to specify thicker copper than the default, which is usually "1 ounce". 2 ounce copper isn't usually that much more money. After that it gets expensive. There is also a limit on how far board houses can go with this. The thickest I've ever heard of is 5 ounce copper.

If this is a one off or low quantity, then leaving the solder mask off the trace and soldering a wire over it is a legitimate thing to do. A #10 copper wire can carry way more current than even a thick PCB trace of reasonable width. Consider how the current has to get onto and off the extra copper wire though. It's easy to solve the bulk conduction problem and forget about the feed points.

Olin Lathrop
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    we used 6 ounce copper on one board, and it wasn't out of the ordinary. If you use >2 ounce copper, you can't use very small traces/spaces on the PCB though. Also it becomes far more difficult to solder through-hole components onto thick copper. – Jason S Sep 03 '11 at 18:23
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Another solution for boards is to make the trace as wide as you can afford (even if it's narrower than calculations, as long as it's not too much so). Make sure the entire trace is NOT masked, then solder-coat the trace, so you have a nice convex bead of solder running the length of the trace. It's probably not the best solution, but I've seen it used in a variety of production electronics, so it can't be that bad (heh).

Doktor J
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If your layout allows it you could place a series of closely spaced filled vias over the length (and width) of the trace. By allowing it I mean that this will of course have its consequences for the bottom layer too. Make the vias as large in diameter as possible, for instance 1mm on a 1.5mm wide trace. Copper filled vias will reduce the trace's resistance best, but they're much more expensive than solder filled vias.

You can also use thicker copper than the standard 35\$\mu\$, like 70\$\mu\$ or even 105\$\mu\$.

stevenvh
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    what about the mechanical consequences of essentially perforating the board? – JustJeff Sep 01 '11 at 11:47
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    @JustJeff - FR4 is very stiff material, you can easily mill a several cm long slot in it without weakening it. So unless you plan to have these traces all over the board + mount a heavy transformer I expect no problems here. I've worked with 0.8mm FR4 and that's stiff enough to carry most components, even with a lot of holes. – stevenvh Sep 01 '11 at 13:49
  • also if you are worried about board warping you can add a cross hatch on the top which prevents that. – quest49 Sep 06 '11 at 23:02
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    Do you have any examples of this? – tyblu May 06 '12 at 22:14
  • @tyblu - not here, but we did it at my previous job to carry 16A from connectors to relays on a relay module for home automation. – stevenvh May 07 '12 at 05:23
  • Could I tell the shop that give me such-micron thick copper-track boards? Are the available ? – Always Confused Jul 04 '16 at 09:02
  • However, this seems to be most appropriate and direct answer to this-question – Always Confused Jul 04 '16 at 09:03
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E-Fab Carries a line of PCB Bus Bars and Stiffeners, our standard products will carry from 16 amps to 128 amps

http://e-fab.com/products/pcb-stiffeners/

Wesley Thornton
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With tinning you can decrease the resistance of the the path by 20% to 70% 1 depending on how thick it's soldered on. If you need just a tad more it seems reasonable.

Soldering a copper wire will bring big gains as a standard PCB is 35µm. In comparison with a 1mm and a 2mm copper wire:

A = h * w = 35µm * 1mm = 35 000 µm²

A = h * w = 35µm * 7mm = 245 000 µm² ~1/7 resistance per length

A = r² * pi = (1mm/2)² * pi = 785 398 µm² ~1/23 resistance per length

A = r² * pi = (2mm/2)² * pi = 3 142 000 µm² ~1/90 resistance per length

[1] EEVBLOG Tinning PCB

P. Koch
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